A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that
1 answer:
Answer:
It travels 1414 feets.
Explanation:
Let's take the length the bullet travels <em>l </em>as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height <em>h</em> it reached, we can calculate <em>l</em> using the <em>sin(α)</em> function:
Replacing:
Solving and roundin to the nearest foot:
You might be interested in
Answer:
t = 2.01 s
Vf = 19.7 m/s
Explanation:
It's know through the International System that the earth's gravity is 9.8 m/s², then we have;
Data:
- Height (h) = 20 m
- Gravity (g) = 9.8 m/s²
- Time (t) = ?
- Final Velocity (Vf) = ?
==================================================================
Time
Use formula:
Replace:
Everything inside the root is solved first. So, we solve the multiplication of the numerator:
It divides:
The square root is performed:
==================================================================
Final Velocity
use formula:
- Vf = g * t
Replace:
- Vf = 9.8 m/s² * 2.01 s
Multiply:
- Vf = 19.7 m/s
==================================================================
How long does it take to reach the ground?
Takes time to reach the ground in <u>2.01 seconds.</u>
How fast does it hit the ground?
Hits the ground with a speed of <u>19.7 meters per seconds.</u>
1) The total mechanical energy of the rock is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
is the mass, 
is the gravitational acceleration and 
is the height.
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is: